Published by:
CGP EDU Academic Team
Published on: September 12, 2026
An elementary particle of mass m and charge +e is projected with velocity v at a much more massive particle of charge ze where $Z > 0.$ What is the closest possible approach of the incident particle
Text Solution
Verified by ExpertsThe correct answer is:
A
Suppose distance of closest approach is r, and according to energy conservation applied for elementary charge. Energy at the time of projection = Energy at the distance of closest approach
⇒ ⇒ $\frac{1}{2}mv^{2} = \frac{1}{4\pi \epsilon_{0}} \cdot \frac{(Ze).e}{r} \Rightarrow r = \frac{Ze^{2}}{2\pi \epsilon_{0} mv^{2}}$
Prepare Smarter with CGP Edu
Get practice questions, solutions, and test series in one place.
Write a Review
Share your experience with this question and solution.
Commentary
Send your comment, doubt, correction, or feedback to admin.
Similar Questions
Explore conceptually related problems
Two equal negative charge – q are fixed at the fixed points (0,a) and $(0,-a)$ on the Y-axis. A pos…
An electric line of force in the xy plane is given by equation $x^{2} + y^{2} = 1$ . A particle wit…
A positively charged ball hangs from a silk thread. We put a positive test charge $\alpha_0$ at a p…
If on the concentric hollow spheres of radii r and $R(>r)$ the charge $QED$ is distributed such tha…
Two equal charges of opposite sign separated by a distance constitute an electric dipole of dipol…
A point charge q is placed at a distance a/2 directly above the center of a square of side a. The e…